Problem: For a function $g$, we are given that $g(-1)=3$ and $g'(-1)=-2$. What's the equation of the tangent line to the graph of $g$ at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y+2=3(x+1)$ (Choice B) B $y+1=-2(x-3)$ (Choice C) C $y+1=3(x+2)$ (Choice D) D $y-3=-2(x+1)$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $g'(-1)$ gives the slope of the tangent line to the graph of $g$ where $x=-1$. We are given that $g'(-1)=-2$, so the slope of the tangent line is $-2$. Furthermore, we are given that $g(-1)=3$, which means the point of intersection of the tangent line and the graph is $(-1,3)$. To summarize, the tangent line has a slope of $-2$ and it passes through the point $(-1,3)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-3&=-2(x-(-1)) \\\\ y-3&=-2(x+1) \end{aligned}$ The equation is $y-3=-2(x+1)$.